Question

Could someone do all the problems and/or check my work?

Answer 1

11. You’ve done it correctly

12. Let x^2=y
y^2+13y+40=0
(y+8)(y+5)=0
y=8, 5
Since y=x^2
x^2=8 x^2=5
x=+/-√5 x= +/-2√2

13. x^4-x^2-x^2-8=0
x^4-2x^2-8=0
let x^2=y
Y^2-2y-8=0
(y-4)(y+2)=0
y=4, -2
Since y=x^2
X^2=4 X^2=-2
X= +/- 2 This wouldn’t be a real solution

14. It’s pretty much the same process, just substitute y in for x^2. If you’re confused feel free to ask and I can do it, or you can put it through Photomath

15. You’re on the right track so I’m just going to continue from where you left off
x^2(4x+5)-4(x+5)=0
(x^2-4)(4x+5)=0
x= +/- 2 4x=5
x=5/4 or 1 1/4

Hope this helped :)

Answer 2

NOTES: To find the intercepts/roots:

• move everything to one side and 0 on the other
• factor the equation
• apply the Zero Product Property (set each factor equal to 0)
• solve for x

11. Answer: x = {-4, 0, 5}

Step-by-step explanation:

$x^3=x^2+20x\\x^3-x^2-20x=0\\x(x^2-x-20)=0\\x(x-5)(x+4)=0\\x=0\quad x-5=0\quad x+4=0\\x=0\quad x=5\quad x=-4\\Solutions:\ x=\{-4, 0, 5\}$

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12. Answer: x = No real solution

Step-by-step explanation:

$x^4+13x^2+40=0\\(x^2+5)(x^2+8)=0\\x^2+5=0\quad \qquad x^2+8=0\\x^2=-5\qquad \qquad x^2=-8\\x=\pm \sqrt{-5}\quad \quad x=\pm \sqrt{-8}\\ \text{Neither solution is a real number}$

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13. Answer: x = {-2, 2}

Step-by-step explanation:

$x^4-x^2=x^2+8\\x^4-2x^2-8=0\\(x^2-4)(x^2+2)=0\\x^2-4=0\quad \qquad x^2+2=0\\x^2=4\qquad \qquad x^2=-2\\x=\pm \sqrt{4}\quad \quad x=\pm \sqrt{-2}\\x=\pm 2\qquad \text{x is not a real solution}\\Solutions: x = \{-2, 2\}$

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14. Answer: $\bold{x=\bigg\{-1, 1, -\dfrac{\sqrt{15}}{3}, \dfrac{\sqrt{15}}{3}\bigg\}}$

Step-by-step explanation:

$3x^4-8x^2+5=0\\3x^4-3x^2-5x^2+5=0\\3x^2(x^2-1)-5(x^2-1)=0\\(3x^2-5)(x^2-1)=0\\3x^2-5=0\qquad x^2-1=0\\x^2=\dfrac{5}{3}\qquad \qquad x^2=1\\\\x=\pm \sqrt{\dfrac{5}{3}}\quad \quad x=\pm \sqrt{1}\\\\x=\pm \dfrac{\sqrt{15}}{3}\qquad x=\pm1\\\\Solutions: x = \bigg\{-1, 1, -\dfrac{\sqrt{15}}{3}, \dfrac{\sqrt{15}}{3}}\bigg\}$

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15. Answer: $\bold{x=\bigg\{-2, 2, -\dfrac{5}{4}\bigg\}}$

Step-by-step explanation:

$4x^3+5x^2-16x-20=0\\3x^4-3x^2-5x^2+5=0\\x^2(4x+5)-4(4x+5)=0\\(x^2-4)(4x+5)=0\\x^2-4=0\qquad 4x+5=0\\x^2=4\qquad \qquad 4x=-5\\\\x=\pm \sqrt{4}\quad \quad x=\dfrac{-5}{4}\\\\x=\pm 2\qquad \quad x=-\dfrac{5}{4}\\\\ Solutions: x = \bigg\{-2, 2, -\dfrac{5}{4}}\bigg\}$